16.01 g of sulfur trioxide was decomposed into 6.41 g of sulfur and 9.60 g of oxygen. How many grams of sulfur trioxide would be needed to produce 28.80 g of oxygen?

48 grams of sulfur trioxide.

Explanation:

The equation of reaction involves the decomposition of 2 moles of sulfur trioxide (SO3) to produce 2 moles of atomic sulfur (S) and 3 moles of molecular oxygen (O2).

From the equation of reaction above,

3 moles of molecular oxygen (96 g) was produced from 2 moles of sulfur trioxide (160 g)

Mass of sulfur trioxide needed to produce 28.8 g of oxygen = 28.8×160/96 = 48 grams